How do you solve #sqrt(3)sin2x+cos2x=0# for 0 to 2pi?

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Answer 1 · 2015-06-15T17:55:31

Solve #sqrt3.sin 2x + cos 2x = 0#

#sin 2x + (1/sqrt3)cos 2x = 0#.

Replace #tan (pi/6) = (1/sqrt3)# by #(sin (pi/6)/cos (pi/6))#

#sin 2x.cos (pi/6) + sin (pi/6).cos 2x = 0#

#sin (2x + pi/6) = 0#

a. #(2x + pi/6) = 0# -> #2x = - pi/6 -> x = - pi/12 #

b. #(2x + pi/6) = pi# -> #2x = pi - pi/6 = (5pi)/6# ->

-># x = 5pi/12#